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3 years ago

15

Electrons and positrons are produced by the nuclear transformations of protons and neutrons known as beta decay. (a) If a proton

transforms into a neutron, is an electron or a positron produced? (b) If a neutron transforms into a proton, is an electron or a positron produced?

1 answer:

Julli [10]3 years ago

5 0

Answer: a) if a proton transforms to a neutron, a positron is produced

B) when an neutron transforms into a proton, an electron is produced

Explanation:

The both are nuclear decay processes which produce a neutrino and tremendous energy. The conversion of protons to neutrons is an energetically difficult process. However, the conversion of neutrons to electrons is commonly called beta decay in nuclear physics.

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Can You compost salad

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Yes, you can compost lettuce and other salad leaves. Be careful! If the salad is covered in lots of dressing, the oils or fats in the dressing may attract rats or other animals to your compost heap.

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3 years ago

The reaction of a quantity of sulfur trioxide with water gives an acid with a volume of 2 l, containing 1.8 mol of the dissolved

777dan777 [17]

Answer :]

A.)Calculate the mass of ammonium sulfate that would be obtained by reacting with ammonia acid.

<em>Correct me if i'm wrong :]</em>

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2 years ago

What is the empirical formula for a compound that is 83.7% carbon and 16.3% hydrogen?

zubka84 [21]

Hello!

We use the amount in grams (mass ratio) based on the composition of the elements, see: (in 100 g solution)

C: 83.7% = 83,7 g
H: 16.3% = 16.3 g

Let us use the above mentioned data (in g) and values will be converted to amount of substance (number of moles) by dividing by molecular mass (g / mol) each of the values, lets see:

$C: \dfrac{83.7\:\diagup\!\!\!\!\!g}{12\:\diagup\!\!\!\!\!g/mol} \approx 6.975\:mol$

$H: \dfrac{16.3\:\diagup\!\!\!\!\!g}{1\:\diagup\!\!\!\!\!g/mol} = 16.3\:mol$

We note that the values found above are not integers, so let's divide these values by the smallest of them, so that the proportion is not changed, let's see:

$C: \dfrac{6.975}{6.975} = 1$

$H: \dfrac{16.3}{6.975} \approx 2.3$

Note: So the ratio in the smallest whole numbers of carbon to hydrogen is 3:7, t<span>hus, the minimum or empirical formula found for the compound will be:
</span>
$\boxed{\boxed{C_3H_7}}\end{array}}\qquad\checkmark$

I hope this helps. =)

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3 years ago

10. Write the formula for the compounds: copper (1) chloride and copper (II) chloride

Natali [406]

Answer:

Copper(II) chloride (CuCl2) reacts with several metals to produce copper metal or copper(I) chloride (CuCl) with oxidation of the other metal.

Explanation:

3 0

3 years ago

A strontium hydroxide solution is prepared by dissolving 10.60 gg of Sr(OH)2Sr(OH)2 in water to make 47.00 mLmL of solution.What

NeTakaya

Answer:

Approximately $1.854\; \rm mol\cdot L^{-1}$ .

Explanation:

Note that both figures in the question come with four significant figures. Therefore, the answer should also be rounded to four significant figures. Intermediate results should have more significant figures than that.

<h3>Formula mass of strontium hydroxide</h3>

Look up the relative atomic mass of $\rm Sr$ , $\rm O$ , and $\rm H$ on a modern periodic table. Keep at least four significant figures in each of these atomic mass data.

$\rm Sr$ : $87.62$ .
$\rm O$ : $15.999$ .
$\rm H$ : $1.008$ .

Calculate the formula mass of $\rm Sr(OH)_2$ :

$M\left(\rm Sr(OH)_2\right) = 87.62 + 2\times (15.999 + 1.008) = 121.634\; \rm g \cdot mol^{-1}$ .

<h3>Number of moles of strontium hydroxide in the solution</h3>

$M\left(\rm Sr(OH)_2\right) =121.634\; \rm g \cdot mol^{-1}$ means that each mole of $\rm Sr(OH)_2$ formula units have a mass of $121.634\; \rm g$ .

The question states that there are $10.60\; \rm g$ of $\rm Sr(OH)_2$ in this solution.

How many moles of $\rm Sr(OH)_2$ formula units would that be?

$\begin{aligned}n\left(\rm Sr(OH)_2\right) &= \frac{m\left(\rm Sr(OH)_2\right)}{M\left(\rm Sr(OH)_2\right)}\\ &= \frac{10.60\; \rm g}{121.634\; \rm g \cdot mol^{-1}} \approx 8.71467\times 10^{-2}\; \rm mol\end{aligned}$ .

<h3>Molarity of this strontium hydroxide solution</h3>

There are $8.71467\times 10^{-2}\; \rm mol$ of $\rm Sr(OH)_2$ formula units in this $47\; \rm mL$ solution. Convert the unit of volume to liter:

$V = 47\; \rm mL = 0.047\; \rm L$ .

The molarity of a solution measures its molar concentration. For this solution:

$\begin{aligned}c\left(\rm Sr(OH)_2\right) &= \frac{n\left(\rm Sr(OH)_2\right)}{V}\\ &= \frac{8.71467\times 10^{-2}\; \rm mol}{0.047\; \rm L} \approx 1.854\; \rm mol \cdot L^{-1}\end{aligned}$ .

(Rounded to four significant figures.)

5 0

3 years ago